Question:

Find the equation of the bisector of the pair of acute angles formed by the lines $\displaystyle L_1$ and $\displaystyle L_2$ given by $\displaystyle 4x-2y-10=0$ and $\displaystyle x-2y-1=0$, respectively.

Attempt:

$\displaystyle L_1 \rightarrow 4x - 2y - 15 = 0$

$\displaystyle A = 4, \ B = -2, \ C=-15 $

$\displaystyle d_1 = \frac{Ax+By+C}{\pm\sqrt{A^2+B^2}} = \frac{4x-2y-15}{-\sqrt{4^2+2^2}} = \frac{4x-2y-15}{-2\sqrt{5}}$

$\displaystyle L_2 \rightarrow x - 2y - 1$

$\displaystyle A = 1, \ B = -2, \ C=-1 $

$\displaystyle d_2 = \frac{Ax+By+C}{\pm\sqrt{A^2+B^2}} = \frac{x-2y-1}{-\sqrt{1^2+2^2}} = \frac{x-2y-1}{-\sqrt{5}}$

Now, I need to simplify the two distance equations to get the equation of the bisector...

$\displaystyle \frac{4x-2y-15}{-2\sqrt{5}} = \frac{x-2y-1}{-\sqrt{5}}$

Here is the problem! I don't know how to simplify with the square roots around , Need Help!